Question

Suppose a researcher is estimating the proportion of trees showing signs of disease in a local forest. They randomly sample 150 trees and find that 25 of them show signs of disease. Calculate the 95% confidence interval for p, the proportion of all trees in the forest showing signs of disease.

Answer 1

`0.167 - 1.96 \sqrt{(0.167(1-0.167))/(150)}=0.107`

`0.167 + 1.96 \sqrt{(0.167(1-0.167))/(150)}=0.227`

And the 95% confidence interval would be given (0.107;0.227).

Explanation:

Information given

`X= 25` number of trees with signs of disease

`n= 150` the sample selected

`\hat p=(25)/(150)= 0.167` the proportion of trees with signs of disease

The confidence interval for the true proportion would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the significance is
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, and the critical value would be

`z_(\alpha/2)=1.96`

And replacing we got:

`0.167 - 1.96 \sqrt{(0.167(1-0.167))/(150)}=0.107`

`0.167 + 1.96 \sqrt{(0.167(1-0.167))/(150)}=0.227`

And the 95% confidence interval would be given (0.107;0.227).